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How to show that $(x+y)^{a} \geq x^{a} + y^{a}$ given that $a \geq 1$ and $x,y \geq 0$? Also, how to prove that the reverse inequality holds when $0 \leq a \leq 1$?

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1 Answer 1

up vote 2 down vote accepted

We deal with $a\ge 1$. If $a=1$ there is nothing to do. So let $a\gt 1$. Fix $y$, and let $f(x)=(x+y)^a-x^a-y^a$.

We have $f(0)=0$, and $f'(x)=a((x+y)^{a-1}-x^{a-1})$. So $f'(x)\ge 0$ if $x\ge 0$. It follows that for fixed $y$, $f(x)$ is non-decreasing.

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oh, thank you very much! Nice proof! – Sleepingip Oct 26 '12 at 14:35

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